GSEB Class 12 Maths Solutions Chapter 10 સદિશ બીજગણિત Exercise 10.2

Question 1. નીચે આપેલા સદિશોનાં માનની ગણતરી કરો:
(i) \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \)
(ii) \( \vec{b} = 2\hat{i} - 7\hat{j} - 3\hat{k} \)
(iii) \( \vec{c}= \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k} \)
Answer:
(i) For the vector \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \), its magnitude is determined as:
\( |\vec{a}| = \sqrt{(1)^2+(1)^2+(1)^2} = \sqrt{1+1+1} = \sqrt{3} \)
So, the magnitude of vector \( \vec{a} \) is \( \sqrt{3} \).
(ii) For the vector \( \vec{b} = 2\hat{i} - 7\hat{j} - 3\hat{k} \), its magnitude is calculated as:
\( |\vec{b}| = \sqrt{(2)^2+(-7)^2+(-3)^2} \)
\( = \sqrt{4+49+9} = \sqrt{62} \)
Thus, the magnitude of vector \( \vec{b} \) is \( \sqrt{62} \).
(iii) For the vector \( \vec{c} = \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k} \), its magnitude is:
\( |\vec{c}| = \sqrt{\left(\frac{1}{\sqrt{3}}\right)^2 + \left(\frac{1}{\sqrt{3}}\right)^2 + \left(\frac{1}{\sqrt{3}}\right)^2} \)
\( = \sqrt{\frac{1}{3} + \frac{1}{3} + \frac{1}{3}} = \sqrt{\frac{3}{3}} = \sqrt{1} = 1 \)
Therefore, the magnitude of vector \( \vec{c} \) is 1.
In simple words: To find the magnitude of a vector, we take the square root of the sum of the squares of its components. This tells us the length of the vector.

Exam Tip: Remember that the magnitude of a vector is always a non-negative scalar quantity. Pay attention to signs when squaring components, as \( (-x)^2 = x^2 \).

Question 2. સમાન માપવાળા બે ભિન્ન સદિશો લખો.
Answer: Let's consider two different vectors, \( \vec{a} \) and \( \vec{b} \), with the same magnitude:
Let \( \vec{a} = \hat{i} + 2\hat{j} + 3\hat{k} \)
The magnitude of \( \vec{a} \) is calculated as:
\( |\vec{a}| = \sqrt{(1)^2 + (2)^2 + (3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \)
Let \( \vec{b} = 3\hat{i} + \hat{j} + 2\hat{k} \)
The magnitude of \( \vec{b} \) is calculated as:
\( |\vec{b}| = \sqrt{(3)^2 + (1)^2 + (2)^2} = \sqrt{9 + 1 + 4} = \sqrt{14} \)
Here, we can observe that \( |\vec{a}| = |\vec{b}| \), but \( \vec{a} \neq \vec{b} \) since their corresponding components are not identical. Many such vectors can be found.
In simple words: You can create many different vectors that have the same length. Just change the order or signs of the numbers inside the vector components, and the length might stay the same even if the direction is different.

Exam Tip: To ensure two vectors are different but have the same magnitude, simply permute the components or change their signs (e.g., \( \langle 1, 2, 3 \rangle \) and \( \langle 2, 1, 3 \rangle \)), as long as the sum of their squares remains the same.

Question 3. જેની દિશા સમાન હોય તેવા બે ભિન્ન સદિશો લખો.
Answer: To find two different vectors with the same direction, one vector must be a scalar multiple of the other, where the scalar is a positive number.
Let's consider two vectors:
\( \vec{a} = 2\hat{i} + 3\hat{j} + 5\hat{k} \)
\( \vec{b} = 4\hat{i} + 6\hat{j} + 10\hat{k} \)
We can see that \( \vec{b} = 2(2\hat{i} + 3\hat{j} + 5\hat{k}) \).
This means \( \vec{b} = 2\vec{a} \). Since 2 is a positive scalar, \( \vec{a} \) and \( \vec{b} \) have the same direction, but they are different vectors because their magnitudes are not equal (e.g., \( |\vec{b}| = 2|\vec{a}| \)). Many such examples can be obtained.
In simple words: Vectors that point in the same direction are parallel. You can make a parallel vector by multiplying the original vector by any positive number; it will still point the same way, but it will be longer or shorter.

Exam Tip: Two vectors \( \vec{u} \) and \( \vec{v} \) have the same direction if \( \vec{v} = k\vec{u} \) for some positive scalar \( k \). If \( k \) is negative, they have opposite directions.

Question 4. સદિશો \( 2\hat{i} + 3\hat{j} \) અને \( x\hat{i} + y\hat{j} \) સમાન થાય તેવી x અને yની કિંમતો શોધો.
Answer: Let the first vector be \( \vec{a} = 2\hat{i} + 3\hat{j} \) and the second vector be \( \vec{b} = x\hat{i} + y\hat{j} \).
For two vectors to be equal, their corresponding components must be identical.
Given that \( \vec{a} = \vec{b} \), we have:
\( 2\hat{i} + 3\hat{j} = x\hat{i} + y\hat{j} \)
By equating the components of \( \hat{i} \) and \( \hat{j} \), we get:
\( x = 2 \)
\( y = 3 \)
Thus, the values of x and y are 2 and 3, respectively, for the vectors to be equal.
In simple words: If two vectors are exactly the same, it means the number in front of the 'i' must match, and the number in front of the 'j' must also match.

Exam Tip: Remember that for two vectors to be equal, both their magnitude and direction must be the same, which implies that their corresponding components must be equal.

Question 5. જે સદિશનું પ્રારંભ બિંદુ (2, 1) અને અંતિમ બિંદુ (–5, 7) હોય, તેના અદિશ અને સદિશ ઘટકો શોધો.
Answer: Let the initial point be A(2, 1) and the terminal point be B(-5, 7).
The position vector of point A is \( \overrightarrow{OA} = 2\hat{i} + \hat{j} \).
The position vector of point B is \( \overrightarrow{OB} = -5\hat{i} + 7\hat{j} \).
The vector \( \overrightarrow{AB} \) is found by subtracting the initial position vector from the terminal position vector:
\( \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} \)
\( = (-5\hat{i} + 7\hat{j}) - (2\hat{i} + \hat{j}) \)
\( = (-5 - 2)\hat{i} + (7 - 1)\hat{j} \)
\( = -7\hat{i} + 6\hat{j} \)
From this vector, the scalar components are the coefficients of \( \hat{i} \) and \( \hat{j} \), which are -7 and 6.
The vector components are the terms involving \( \hat{i} \) and \( \hat{j} \), which are \( -7\hat{i} \) and \( 6\hat{j} \).
In simple words: To find a vector that goes from one point to another, subtract the coordinates of the start point from the end point. The resulting numbers are the scalar components, and adding the direction symbols makes them vector components.

Exam Tip: A vector from point A to point B is always \( \overrightarrow{B} - \overrightarrow{A} \). Scalar components are just the numbers, while vector components include the unit vectors (\( \hat{i}, \hat{j}, \hat{k} \)).

Question 6. સદિશો \( \vec{a} = \hat{i} – 2\hat{j} + \hat{k} \), \( \vec{b} = -2\hat{i} + 4\hat{j} + 5\hat{k} \) અને \( \vec{c} = \hat{i} – 6\hat{j} – 7\hat{k} \) નો સરવાળો શોધો.
Answer: We are given the following vectors:
\( \vec{a} = \hat{i} – 2\hat{j} + \hat{k} \)
\( \vec{b} = -2\hat{i} + 4\hat{j} + 5\hat{k} \)
\( \vec{c} = \hat{i} – 6\hat{j} – 7\hat{k} \)
To find their sum, we add the corresponding components (coefficients of \( \hat{i}, \hat{j}, \hat{k} \)):
\( \vec{a}+\vec{b}+\vec{c} = (\hat{i} – 2\hat{j} + \hat{k}) + (-2\hat{i} + 4\hat{j} + 5\hat{k}) + (\hat{i} – 6\hat{j} – 7\hat{k}) \)
\( = (1 - 2 + 1)\hat{i} + (-2 + 4 - 6)\hat{j} + (1 + 5 - 7)\hat{k} \)
\( = 0\hat{i} - 4\hat{j} - \hat{k} \)
\( = -4\hat{j} - \hat{k} \)
The sum of the vectors is \( -4\hat{j} - \hat{k} \).
In simple words: To add vectors, you just add up all the 'i' parts together, all the 'j' parts together, and all the 'k' parts together. Each direction is added separately.

Exam Tip: When adding vectors, be careful with the signs of the components. Summing coefficients of each unit vector separately helps prevent errors.

Question 7. સદિશો \( \vec{a} = \hat{i} + \hat{j} + 2\hat{k} \)ની દિશામાં એકમ સદિશ શોધો.
Answer: Let the given vector be \( \vec{a} = \hat{i} + \hat{j} + 2\hat{k} \).
First, we need to calculate the magnitude of vector \( \vec{a} \):
\( |\vec{a}| = \sqrt{(1)^2 + (1)^2 + (2)^2} \)
\( = \sqrt{1 + 1 + 4} = \sqrt{6} \)
A unit vector in the direction of \( \vec{a} \) is found by dividing the vector by its magnitude:
\( \hat{a} = \frac{\vec{a}}{|\vec{a}|} \)
\( = \frac{1}{\sqrt{6}}(\hat{i} + \hat{j} + 2\hat{k}) \)
\( = \frac{1}{\sqrt{6}}\hat{i} + \frac{1}{\sqrt{6}}\hat{j} + \frac{2}{\sqrt{6}}\hat{k} \)
This is the unit vector in the direction of \( \vec{a} \).
In simple words: A unit vector is like a tiny arrow pointing in the same direction as your main vector, but its length is always exactly one. You find it by dividing each part of your vector by its total length.

Exam Tip: A unit vector always has a magnitude of 1. Remember to divide each component of the original vector by its magnitude to obtain the unit vector.

Question 8. જો P અને Q અનુક્રમે બિંદુઓ (1, 2, 3) અને (4, 5, 6) હોય તો PQની દિશામાં એકમ સદિશ શોધો.
Answer: Given points P(1, 2, 3) and Q(4, 5, 6).
The position vector of point P is \( \overrightarrow{OP} = \hat{i} + 2\hat{j} + 3\hat{k} \).
The position vector of point Q is \( \overrightarrow{OQ} = 4\hat{i} + 5\hat{j} + 6\hat{k} \).
First, we determine the vector \( \overrightarrow{PQ} \):
\( \overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} \)
\( = (4\hat{i} + 5\hat{j} + 6\hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k}) \)
\( = (4 - 1)\hat{i} + (5 - 2)\hat{j} + (6 - 3)\hat{k} \)
\( = 3\hat{i} + 3\hat{j} + 3\hat{k} \)
Next, we calculate the magnitude of \( \overrightarrow{PQ} \):
\( |\overrightarrow{PQ}| = \sqrt{(3)^2 + (3)^2 + (3)^2} \)
\( = \sqrt{9 + 9 + 9} = \sqrt{27} \)
\( = \sqrt{9 \times 3} = 3\sqrt{3} \)
Finally, we find the unit vector in the direction of \( \overrightarrow{PQ} \):
\( \widehat{PQ} = \frac{\overrightarrow{PQ}}{|\overrightarrow{PQ}|} \)
\( = \frac{1}{3\sqrt{3}}(3\hat{i} + 3\hat{j} + 3\hat{k}) \)
\( = \frac{3}{3\sqrt{3}}\hat{i} + \frac{3}{3\sqrt{3}}\hat{j} + \frac{3}{3\sqrt{3}}\hat{k} \)
\( = \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k} \)
This is the unit vector in the direction of PQ.
In simple words: To find the unit vector between two points, first figure out the vector itself by subtracting the start point from the end point. Then, divide this new vector by its length to get the unit vector.

Exam Tip: Always remember that \( \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} \). Simplify the square root of the magnitude for a cleaner final answer, such as \( \sqrt{27} = 3\sqrt{3} \).

Question 9. આપેલ સદિશો \( \vec{a} = 2\hat{i} – \hat{j} + 2\hat{k} \) અને \( \vec{b} = – \hat{i} + \hat{j} – \hat{k} \) હોય, તો સદિશ \( \vec{a} + \vec{b} \) ની દિશામાં એકમ સદિશ શોધો.
Answer: We are given two vectors:
\( \vec{a} = 2\hat{i} – \hat{j} + 2\hat{k} \)
\( \vec{b} = – \hat{i} + \hat{j} – \hat{k} \)
First, we need to find the sum of these two vectors, \( \vec{a} + \vec{b} \):
\( \vec{a}+\vec{b} = (2\hat{i} – \hat{j} + 2\hat{k}) + (– \hat{i} + \hat{j} – \hat{k}) \)
\( = (2 - 1)\hat{i} + (-1 + 1)\hat{j} + (2 - 1)\hat{k} \)
\( = 1\hat{i} + 0\hat{j} + 1\hat{k} \)
\( = \hat{i} + \hat{k} \)
Next, we calculate the magnitude of the resultant vector \( \vec{a} + \vec{b} \):
\( |\vec{a}+\vec{b}| = \sqrt{(1)^2 + (0)^2 + (1)^2} = \sqrt{1 + 0 + 1} = \sqrt{2} \)
Finally, the unit vector in the direction of \( \vec{a} + \vec{b} \) is:
\( \widehat{(\vec{a}+\vec{b})} = \frac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|} \)
\( = \frac{1}{\sqrt{2}}(\hat{i} + \hat{k}) \)
\( = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{k} \)
In simple words: To get a unit vector for the sum of two vectors, first add them up. Then, find the length of this new sum vector. Finally, divide the sum vector by its length.

Exam Tip: When calculating the sum of vectors, ensure you combine coefficients for each unit vector separately. A missing \( \hat{j} \) component implies its coefficient is 0, which should be included in magnitude calculations.

Question 10. \( 5\hat{i} - \hat{j} + 2\hat{k} \) સદિશની દિશામાં 8 એકમ માનવાળો સદિશ શોધો.
Answer: Let the given vector be \( \vec{a} = 5\hat{i} - \hat{j} + 2\hat{k} \).
First, we need to find the unit vector in the direction of \( \vec{a} \). To do this, we calculate its magnitude:
\( |\vec{a}| = \sqrt{(5)^2 + (-1)^2 + (2)^2} \)
\( = \sqrt{25 + 1 + 4} = \sqrt{30} \)
The unit vector \( \hat{a} \) in the direction of \( \vec{a} \) is:
\( \hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{1}{\sqrt{30}}(5\hat{i} - \hat{j} + 2\hat{k}) \)
Now, to find a vector with a magnitude of 8 units in the same direction as \( \vec{a} \), we multiply the unit vector \( \hat{a} \) by 8:
Vector with magnitude 8 \( = 8\hat{a} \)
\( = 8 \times \frac{1}{\sqrt{30}}(5\hat{i} - \hat{j} + 2\hat{k}) \)
\( = \frac{40}{\sqrt{30}}\hat{i} - \frac{8}{\sqrt{30}}\hat{j} + \frac{16}{\sqrt{30}}\hat{k} \)
This is the vector with magnitude 8 in the specified direction.
In simple words: To get a vector with a specific length in a certain direction, first make the original vector into a unit vector (length 1), then multiply that unit vector by the desired length.

Exam Tip: To create a vector of specific magnitude M in the direction of \( \vec{v} \), calculate the unit vector \( \hat{v} = \frac{\vec{v}}{|\vec{v}|} \) and then multiply it by M: \( M\hat{v} \).

Question 11. દર્શાવો કે સદિશો \( 2\hat{i} – 3\hat{j} + 4\hat{k} \) અને \( -4\hat{i} + 6\hat{j} – 8\hat{k} \) સમરેખ છે.
Answer: Let the two given vectors be \( \vec{A} \) and \( \vec{B} \).
\( \vec{A} = 2\hat{i} – 3\hat{j} + 4\hat{k} \)
\( \vec{B} = -4\hat{i} + 6\hat{j} – 8\hat{k} \)
We can factor out a scalar from vector \( \vec{B} \):
\( \vec{B} = -2(2\hat{i} – 3\hat{j} + 4\hat{k}) \)
Substituting \( \vec{A} \) into this equation, we get:
\( \vec{B} = -2\vec{A} \)
Since \( \vec{B} \) can be expressed as a scalar multiple of \( \vec{A} \) (where the scalar is -2), the two vectors \( \vec{A} \) and \( \vec{B} \) are parallel. If two vectors are parallel and share a common initial point (or can be represented as position vectors from the origin), they are considered collinear. Therefore, the vectors \( \vec{A} \) and \( \vec{B} \) are collinear.
In simple words: Two vectors are collinear if one is just a scaled version of the other. If you can multiply a vector by a number (positive or negative) to get the second vector, they lie on the same line.

Exam Tip: Two vectors \( \vec{u} \) and \( \vec{v} \) are collinear if and only if \( \vec{v} = k\vec{u} \) for some scalar \( k \). If \( k>0 \), they are in the same direction; if \( k<0 \), they are in opposite directions, but still collinear.

Question 12. સદિશ \( \hat{i} + 2\hat{j} + 3\hat{k} \) ના દિક્કોસાઇન શોધો.
Answer: Let the given vector be \( \vec{r} = \hat{i} + 2\hat{j} + 3\hat{k} \).
The components of the vector are \( x = 1, y = 2, z = 3 \).
First, we calculate the magnitude of the vector \( \vec{r} \):
\( |\vec{r}| = \sqrt{(1)^2 + (2)^2 + (3)^2} \)
\( = \sqrt{1 + 4 + 9} = \sqrt{14} \)
The direction cosines (l, m, n) are given by the formulas:
\( l = \frac{x}{|\vec{r}|} = \frac{1}{\sqrt{14}} \)
\( m = \frac{y}{|\vec{r}|} = \frac{2}{\sqrt{14}} \)
\( n = \frac{z}{|\vec{r}|} = \frac{3}{\sqrt{14}} \)
Thus, the direction cosines of the vector \( \hat{i} + 2\hat{j} + 3\hat{k} \) are \( \frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}} \).
In simple words: Direction cosines tell you how much a vector aligns with each of the x, y, and z axes. You find them by dividing each vector component by the vector's total length.

Exam Tip: The sum of the squares of the direction cosines of any vector is always 1, i.e., \( l^2 + m^2 + n^2 = 1 \). This can be used as a quick check for your answer.

Question 13. જે સદિશ બિંદુઓ A (1, 2, –3) અને B (–1, –2, 1)ને Aથી B તરફની દિશામાં જોડતો હોય તે સદિશના દિક્કોસાઇન શોધો.
Answer: Given initial point A(1, 2, -3) and terminal point B(-1, -2, 1).
The position vector of A is \( \overrightarrow{OA} = \hat{i} + 2\hat{j} – 3\hat{k} \).
The position vector of B is \( \overrightarrow{OB} = -\hat{i} – 2\hat{j} + \hat{k} \).
First, we determine the vector \( \overrightarrow{AB} \):
\( \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} \)
\( = (-\hat{i} – 2\hat{j} + \hat{k}) - (\hat{i} + 2\hat{j} – 3\hat{k}) \)
\( = (-1 - 1)\hat{i} + (-2 - 2)\hat{j} + (1 - (-3))\hat{k} \)
\( = -2\hat{i} – 4\hat{j} + 4\hat{k} \)
Next, we calculate the magnitude of \( \overrightarrow{AB} \):
\( |\overrightarrow{AB}| = \sqrt{(-2)^2 + (-4)^2 + (4)^2} \)
\( = \sqrt{4 + 16 + 16} = \sqrt{36} = 6 \)
The components of \( \overrightarrow{AB} \) are \( x = -2, y = -4, z = 4 \).
The direction cosines are:
\( l = \frac{x}{|\overrightarrow{AB}|} = \frac{-2}{6} = -\frac{1}{3} \)
\( m = \frac{y}{|\overrightarrow{AB}|} = \frac{-4}{6} = -\frac{2}{3} \)
\( n = \frac{z}{|\overrightarrow{AB}|} = \frac{4}{6} = \frac{2}{3} \)
Therefore, the direction cosines of the vector \( \overrightarrow{AB} \) are \( -\frac{1}{3}, -\frac{2}{3}, \frac{2}{3} \).
In simple words: To find the direction cosines of a vector between two points, first find the vector by subtracting the start point from the end point. Then, divide each part of this new vector by its total length to get the cosines.

Exam Tip: Pay close attention to the order of points when forming the vector (A to B means \( \overrightarrow{B} - \overrightarrow{A} \)). Also, simplify fractions for direction cosines to their lowest terms.

Question 14. સદિશ \( \hat{i} + \hat{j} + \hat{k} \) એ અક્ષો OX, OY અને OZ સાથે સમાન ખૂણા બનાવે છે.
Answer: Let the given vector be \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \).
The components of the vector are \( x = 1, y = 1, z = 1 \).
First, we calculate the magnitude of the vector \( \vec{a} \):
\( |\vec{a}| = \sqrt{(1)^2 + (1)^2 + (1)^2} \)
\( = \sqrt{1 + 1 + 1} = \sqrt{3} \)
Let \( \alpha, \beta, \gamma \) be the angles that the vector \( \vec{a} \) makes with the positive x-axis, y-axis, and z-axis, respectively. The cosines of these angles are the direction cosines:
\( \cos \alpha = \frac{x}{|\vec{a}|} = \frac{1}{\sqrt{3}} \)
\( \cos \beta = \frac{y}{|\vec{a}|} = \frac{1}{\sqrt{3}} \)
\( \cos \gamma = \frac{z}{|\vec{a}|} = \frac{1}{\sqrt{3}} \)
Since \( \cos \alpha = \cos \beta = \cos \gamma = \frac{1}{\sqrt{3}} \), it means that the angles themselves are equal: \( \alpha = \beta = \gamma \).
Therefore, the vector \( \hat{i} + \hat{j} + \hat{k} \) makes equal angles with the coordinate axes OX, OY, and OZ.
In simple words: A vector forms equal angles with the x, y, and z axes if all its components are the same. This means it points equally in all three directions.

Exam Tip: If the direction cosines are equal, then the angles the vector makes with the axes are also equal. This property is common for vectors like \( \hat{i} + \hat{j} + \hat{k} \).

Question 15. બિંદુ R એ બિંદુઓ P અને Qને જોડતા રેખાખંડનું 2:1 ગુણોત્તરમાં (i) અંતઃ (ii) બહિર્વિભાજન કરે છે. P અને Qના સ્થાનસદિશો અનુક્રમે \( \hat{i} + 2\hat{j} – \hat{k} \) અને \( -\hat{i} + \hat{j} + \hat{k} \) છે, તો બિંદુ Rનો સ્થાનસદિશ શોધો.
Answer: Given position vectors of P and Q:
\( \overrightarrow{OP} = \hat{i} + 2\hat{j} – \hat{k} \)
\( \overrightarrow{OQ} = -\hat{i} + \hat{j} + \hat{k} \)
The ratio of division is m:n = 2:1.

(i) For internal division, the position vector of point R is given by the section formula:
\( \overrightarrow{OR} = \frac{m\overrightarrow{OQ} + n\overrightarrow{OP}}{m+n} \)
Substituting the given values:
\( \overrightarrow{OR} = \frac{2(-\hat{i} + \hat{j} + \hat{k}) + 1(\hat{i} + 2\hat{j} – \hat{k})}{2+1} \)
\( = \frac{-2\hat{i} + 2\hat{j} + 2\hat{k} + \hat{i} + 2\hat{j} – \hat{k}}{3} \)
\( = \frac{(-2+1)\hat{i} + (2+2)\hat{j} + (2-1)\hat{k}}{3} \)
\( = \frac{-\hat{i} + 4\hat{j} + \hat{k}}{3} \)
\( = -\frac{1}{3}\hat{i} + \frac{4}{3}\hat{j} + \frac{1}{3}\hat{k} \)
So, the position vector of R when dividing internally is \( -\frac{1}{3}\hat{i} + \frac{4}{3}\hat{j} + \frac{1}{3}\hat{k} \).

(ii) For external division, the position vector of point R is given by the section formula:
\( \overrightarrow{OR} = \frac{m\overrightarrow{OQ} - n\overrightarrow{OP}}{m-n} \)
Substituting the given values:
\( \overrightarrow{OR} = \frac{2(-\hat{i} + \hat{j} + \hat{k}) - 1(\hat{i} + 2\hat{j} – \hat{k})}{2-1} \)
\( = \frac{-2\hat{i} + 2\hat{j} + 2\hat{k} - \hat{i} - 2\hat{j} + \hat{k}}{1} \)
\( = (-2-1)\hat{i} + (2-2)\hat{j} + (2+1)\hat{k} \)
\( = -3\hat{i} + 0\hat{j} + 3\hat{k} \)
\( = -3\hat{i} + 3\hat{k} \)
So, the position vector of R when dividing externally is \( -3\hat{i} + 3\hat{k} \).
In simple words: When a point divides a line segment, it can do so from the inside (internal division) or from the outside (external division). We use different formulas to find its exact location vector, based on the given ratio.

Exam Tip: Remember the internal division formula \( \overrightarrow{OR} = \frac{m\overrightarrow{OQ} + n\overrightarrow{OP}}{m+n} \) and the external division formula \( \overrightarrow{OR} = \frac{m\overrightarrow{OQ} - n\overrightarrow{OP}}{m-n} \). Pay attention to the signs in the external division formula.

Question 16. બિંદુઓ P (2, 3, 4) અને Q (4, 1, –2)ને જોડતા રેખાખંડના મધ્યબિંદુનો સ્થાનસદિશ શોધો.
Answer: Given points P(2, 3, 4) and Q(4, 1, -2).
The position vector of P is \( \overrightarrow{OP} = 2\hat{i} + 3\hat{j} + 4\hat{k} \).
The position vector of Q is \( \overrightarrow{OQ} = 4\hat{i} + \hat{j} – 2\hat{k} \).
Let R be the midpoint of the line segment PQ. The position vector of the midpoint is given by:
\( \overrightarrow{OR} = \frac{\overrightarrow{OP} + \overrightarrow{OQ}}{2} \)
Substituting the given position vectors:
\( \overrightarrow{OR} = \frac{(2\hat{i} + 3\hat{j} + 4\hat{k}) + (4\hat{i} + \hat{j} – 2\hat{k})}{2} \)
Combine the corresponding components:
\( = \frac{(2+4)\hat{i} + (3+1)\hat{j} + (4-2)\hat{k}}{2} \)
\( = \frac{6\hat{i} + 4\hat{j} + 2\hat{k}}{2} \)
Divide each component by 2:
\( = 3\hat{i} + 2\hat{j} + \hat{k} \)
Therefore, the position vector of the midpoint R is \( 3\hat{i} + 2\hat{j} + \hat{k} \).
In simple words: To find the middle point of a line segment, you just average the position vectors of its two end points. Add them up, then divide by two.

Exam Tip: The midpoint formula is a special case of the internal division formula where the ratio is 1:1. Remember to divide *each* component by 2.

Question 17. સાબિત કરો કે બિંદુઓ A, B અને Cના સ્થાનસદિશો અનુક્રમે \( \vec{a} = 3\hat{i} - 4\hat{j} - 4\hat{k} \), \( \vec{b} = 2\hat{i} – \hat{j} + \hat{k} \), \( \vec{c} = \hat{i} – 3\hat{j} – 5\hat{k} \) હોય, તો તે કાટકોણ ત્રિકોણ રચે છે.
Answer: Let the position vectors of points A, B, and C be:
\( \overrightarrow{OA} = 3\hat{i} – 4\hat{j} – 4\hat{k} \)
\( \overrightarrow{OB} = 2\hat{i} – \hat{j} + \hat{k} \)
\( \overrightarrow{OC} = \hat{i} – 3\hat{j} – 5\hat{k} \)
To prove that these points form a right-angled triangle, we need to calculate the vectors representing the sides of the triangle and their magnitudes. Then, we can check if the Pythagorean theorem holds.

First, find the vectors for the sides:
\( \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (2\hat{i} – \hat{j} + \hat{k}) - (3\hat{i} – 4\hat{j} – 4\hat{k}) \)
\( = (2-3)\hat{i} + (-1-(-4))\hat{j} + (1-(-4))\hat{k} \)
\( = -\hat{i} + 3\hat{j} + 5\hat{k} \)

\( \overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = (\hat{i} – 3\hat{j} – 5\hat{k}) - (2\hat{i} – \hat{j} + \hat{k}) \)
\( = (1-2)\hat{i} + (-3-(-1))\hat{j} + (-5-1)\hat{k} \)
\( = -\hat{i} – 2\hat{j} – 6\hat{k} \)

\( \overrightarrow{CA} = \overrightarrow{OA} - \overrightarrow{OC} = (3\hat{i} – 4\hat{j} – 4\hat{k}) - (\hat{i} – 3\hat{j} – 5\hat{k}) \)
\( = (3-1)\hat{i} + (-4-(-3))\hat{j} + (-4-(-5))\hat{k} \)
\( = 2\hat{i} – \hat{j} + \hat{k} \)

Next, calculate the square of the magnitudes of these vectors:
\( |\overrightarrow{AB}|^2 = (-1)^2 + (3)^2 + (5)^2 = 1 + 9 + 25 = 35 \)
\( |\overrightarrow{BC}|^2 = (-1)^2 + (-2)^2 + (-6)^2 = 1 + 4 + 36 = 41 \)
\( |\overrightarrow{CA}|^2 = (2)^2 + (-1)^2 + (1)^2 = 4 + 1 + 1 = 6 \)

Now, we check if the Pythagorean theorem holds for any combination of these squared magnitudes:
Consider \( |\overrightarrow{AB}|^2 + |\overrightarrow{CA}|^2 = 35 + 6 = 41 \)
We observe that \( |\overrightarrow{AB}|^2 + |\overrightarrow{CA}|^2 = |\overrightarrow{BC}|^2 \).
By the converse of the Pythagorean theorem, since the sum of the squares of two sides equals the square of the third side, the triangle ABC is a right-angled triangle. The right angle is at vertex A, opposite the hypotenuse BC.
In simple words: To prove a triangle is right-angled using vectors, calculate the length of each side. Then, square those lengths. If the sum of the squares of two shorter sides equals the square of the longest side, it's a right triangle.

Exam Tip: For right-angled triangle problems, always calculate \( |\text{side}|^2 \) for all three sides. The longest side will be the hypotenuse, and its square should equal the sum of the squares of the other two sides. The angle opposite the longest side is the right angle.

Question 18. પ્રશ્નો 18 તથા 19માં વિધાન સાચું બને તે રીતે આપેલ વિકલ્પોમાંથી યોગ્ય વિકલ્પ પસંદ કરો : ઓ આકૃતિ માટે નીચેનામાંથી કયાં વિધાનો સત્ય નથી :
(A) \( \overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=\overrightarrow{0} \)
(B) \( \overrightarrow{AB}+\overrightarrow{BC}-\overrightarrow{AC}=\overrightarrow{0} \)
(C) \( \overrightarrow{AB}+\overrightarrow{BC}-\overrightarrow{CA}=\overrightarrow{0} \)
(D) \( \overrightarrow{AB}-\overrightarrow{CB}+\overrightarrow{CA}=\overrightarrow{0} \)
Answer: (C) \( \overrightarrow{AB}+\overrightarrow{BC}-\overrightarrow{CA}=\overrightarrow{0} \)
In simple words: For a triangle, when you add vectors that go around the sides in a loop (like A to B, B to C, then C back to A), the total sum is zero. Any statement that doesn't follow this basic rule, or its variations, is incorrect.

Exam Tip: Remember the triangle law of vector addition: \( \overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC} \). Also, recall that \( \overrightarrow{PQ} = -\overrightarrow{QP} \). Use these fundamental rules to test each option.

Question 19. જો \( \vec{a} \) અને \( \vec{b} \), બે સમરેખ સદિશો હોય, તો નીચે આપેલાં પૈકી કયાં વિધાનો અસત્ય છે :
(A) કોઈક અદિશ \( \lambda \) માટે, \( \vec{b} = \lambda\vec{a} \)
(B) \( \vec{a} = \vec{b} \)
(C) \( \vec{a} \) અને \( \vec{b} \) ના અનુરૂપ ઘટકો પ્રમાણમાં નથી.
(D) બંને સદિશો \( \vec{a} \) અને \( \vec{b} \) ની દિશા સમાન છે, પરંતુ માન ભિન્ન છે.
Answer: (B), (C) અને (D) અસત્ય છે.
In simple words: Collinear vectors mean they lie on the same line, but they don't have to be identical, point in the same direction, or have matching components in fixed proportions. We look for statements that wrongly describe these core properties.

Exam Tip: Collinear vectors are vectors that are parallel to the same line. This implies that one vector can be expressed as a scalar multiple of the other (\( \vec{b} = \lambda \vec{a} \)). This relationship means their components are proportional. Their directions can be the same (if \( \lambda > 0 \)) or opposite (if \( \lambda < 0 \)), and their magnitudes can be equal or different.